# The snail and the 10-metre well: visual puzzle | SwaVid

A snail is at the bottom of a 10-metre well. Every day it climbs 3 metres, and every night it slips down 2 metres. On which day does it get out?

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# The snail and the 10-metre well

## The snail gets out on Day 8.

### Find the gain after a complete cycle

### Track the first seven full cycles

### Treat the final day differently

SwaVid visual puzzle

A snail is at the bottom of a 10-metre well. Every day it climbs 3 metres, and every night it slips down 2 metres. On which day does it get out?

Detailed solution

A complete day-and-night cycle gains 1 metre, but there is no night-time slip after the snail reaches the top.

During the day the snail gains 3 metres. During the night it loses 2 metres.

After seven days and seven nights, the snail has made seven net gains of 1 metre and starts Day 8 at 7 metres.

On Day 8 it climbs another 3 metres, reaches 10 metres, and leaves the well. It does not remain in the well for an eighth night, so it never makes that final 2-metre slide.

Dividing 10 by the 1-metre net gain gives 10 days, but that incorrectly adds a night-time slide after the snail has already escaped.

- 1 Find the gain after a complete cycle During the day the snail gains 3 metres. During the night it loses 2 metres. 3 m − 2 m = 1 m net gain
- 2 Track the first seven full cycles After seven days and seven nights, the snail has made seven net gains of 1 metre and starts Day 8 at 7 metres. 7 × 1 m = 7 m
- 3 Treat the final day differently On Day 8 it climbs another 3 metres, reaches 10 metres, and leaves the well. It does not remain in the well for an eighth night, so it never makes that final 2-metre slide. 7 m + 3 m = 10 m

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